Calculate the mass of anhydrous `Ca_3(PO_4)_2` present in 100 ml of 0.25 M solution.

Options

A. 15.45g
B. 7.75g
C. 12.75g
D. 14g

Answer

A, B, C

Explanation

`(A)` lonisation energy `= 13.6 xx 16 eV = 217.6 eV`
`(B) PE = -27.2 xx (4^(2))/(2^(2)) = -108.8e V`
`(C) (f_(2))_(Be^(3+))/(f_(1))_(H) = ((4^(2))/(2^(3)))/(1^(2)/(1^(3))) = 2`
The frequency of revolution of electron in `2^(nd)` orbit is doble of the frequency of revolution of electron in ground state of `H -` atom.
`(D) 2^(nd)` excitation potential `= 12.09 xx 16 = 193.4 V`